We can use the equation for the position, which is $\Delta x = x - x_0 = v_0 t + \frac{1}{2} a t^2$. Since the man starts from rest, $v_0$ is zero. So, $\Delta x = \frac{1}{2} a t^2 = \frac{1}{2} \times 6 \, \mathrm{m/s^2} \times (1.8 \, \mathrm{s})^2 = 9.72 \,
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